Problem Solving

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

This question has been answered before by Stuart before but I really did not understand it. Could someone please help me understand?

The approach I did not understand is here
https://www.beatthegmat.com/gmat-prep-ch ... 16793.html

noone?

Hi,
Let r - rate of reaction
a - concentration of A
b - concentration of B
k - constant of proportionality
So, it is given that r = (ka^2)/b
the concentration of chemical B is increased by 100 percent. So, it becomes 2b
rate remains unchanged. So, r remains same. Let c be the new concentration of A.
So, r = kc^2/2b
ka^2/b = kc^2/2b =>c^2 = 2a^2
So, c = a(sqrt2) = 1.41a
So, concentration of A is increased by [(1.4a - a)/a]*100 = 40%

Hence, D

Let's start with A = 10 and B = 100, so the original ratio is 10^2/100 = 1.

If we double B, we have 10^2/200.

Now let's backsolve:

If we increase A by 50% (choice e), the new ratio is 15^2/200 = 225/200.

If we increase A by 40% (choice d), the new ratio is 14^2/200 = 196/200.

The 40% increase is much closer to our 1:1 ratio - choose (D).